Let 729, 81, 9, 1, ... be a sequence and denote the product of the first terms of this sequence. If and , then is equal to: [2026]

Question

Let 729, 81, 9, 1, ... be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence.

If $2 \sum_{n = 1}^{40} {(P_{n})}^{\frac{1}{n}} = \frac{3^{α} - 1}{3^{β}}$ and $\gcd (α, β) = 1$ , then $α + β$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$P_{n} = 729 \cdot 81 \cdot 9 \dots (n terms)$

$= 3^{6} \cdot 3^{4} \cdot 3^{2} \dots 3^{- 2 n + 8}$

$P_{n} = 3^{6 + 4 + 2 + \dots + (- 2 n + 8)} = 3^{n (7 - n)}$

$P_{n}^{1 / n} = 3^{7 - n}$

$\Rightarrow \sum_{n = 1}^{40} {(P_{n})}^{1 / n} = 3^{6} + 3^{5} + \dots (40 terms)$

$= 3^{6} [\frac{1 - {(\frac{1}{3})}^{40}}{1 - \frac{1}{3}}]$

$= 3^{6} \frac{[3^{40} - 1] \times 3^{1}}{3^{40} \times 2}$

$\sum {(P_{n})}^{1 / n} = \frac{(3^{40} - 1)}{2 \times 3^{33}},$

$α = 40, β = 33$

$α + β = 73$

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